Convergence of $\sum_{n=1}^\infty 2^n\sin\frac{1}{3^nz}$ The problem is:
prove $\sum_{n=1}^\infty 2^n\sin\frac{1}{3^nz}$ converges absolutely for all $z\neq 0$, but does not converge uniformly near $z=0$.
Proof:
for all $z\neq 0$
$$\left|2^n\sin\frac{1}{3^nz}-2^n\frac{1}{3^nz}\right|
= \left|
2^n\left(
    -\frac{(\frac{1}{3^nz})^3}{3!}+\frac{(\frac{1}{3^nz})^5}{5!}\dots
\right)
\right|, \ \ \ \ (1)$$
$\exists~ N$ such that when n>N, $\frac{1/z}{3^n}<1$, so that (1) is less than
$$\left|
2^n\left(
    \frac{|\frac{1}{3^nz}|^3}{3!}+\frac{|\frac{1}{3^nz}|^5}{5!}\dots
\right)
\right|
<\left|
\frac{2^n}{3^nz}\left(
    \frac{|\frac{1}{3^nz}|^2}{1-|\frac{1}{3^nz}|^2}
\right)
\right|
<\frac{2^n}{3^n}\left|
\frac{1}{z}\left(
    \frac{|\frac{1}{3^nz}|^2}{1-|\frac{1}{3^nz}|^2}
\right)
\right|,
$$
$\forall~ \epsilon, \exists~ N_1>-\log(\epsilon^{1/2} z^{3/2})$, such that when $n>N_2=\max\{N, N_1\}$, $|\frac{1}{z}||\frac{1}{3^nz}|^2<\epsilon$, and so (1) is less $\frac{2^n}{3^n}\epsilon$.
Therefore, we have $N_2(\epsilon)$ satisfying that $\forall~ p,$
$$\left|\sum_{n=N_2}^{N_2+p} 2^n\sin\frac{1}{3^nz}-\sum_{n=N_2}^{N_2+p} 2^n\frac{1}{3^nz}\right|
<\sum_{n=N_2}^{N_2+p} \left|2^n\sin\frac{1}{3^nz}-2^n\frac{1}{3^nz}\right|\ \ \ \ (2)\\
<\sum_{n=N_2}^{N_2+p}\frac{2^n}{3^n}\epsilon
\leq 2\epsilon,
$$
and so $\sum_{n=1}^\infty 2^n\sin\frac{1}{3^nz}$ converges absolutely. (A step seems to be missing. One should, instead of $\sum_{n=N_2}^{N_2+p}\frac{2^n}{3^n}$, use something like 2/z (plus a constant), which is the limit of the former.)$\blacksquare$
A possibly trivial question is whether it is proper to prove this way: given n sufficiently large, $u_n<f(n)\epsilon$ (different from that $u_n<\epsilon$, or that $u_n/f(n)<\epsilon$ and so $u_n<f(n)\epsilon$).

There are other questions that I may post somewhere else.

The following is a proof that the series doesn't converge uniformly. It's unnecessarily for answering my questions, but I put it here for completeness of proof.
Proof:
However large $N_2$, we can find $n_0$>$N_2$ and $z=\frac{2}{\pi 3^{n_0}}$ such that $2^{n_0}\sin\frac{1}{3^{n_0}z}=2^{n_0}$, and so (say the limit function is $f(z)=\frac{2}{z}+C$, where $C$ is a constant; I suddenly realize C seems also to be a function of z, that could cause some issues),
$$\left|\sum_{n=1}^{\infty} 2^n\sin\frac{1}{3^nz}-f(z)\right|
>\left|\sum_{n=N_2}^{N_2+p} 2^n\sin\frac{1}{3^nz}-\sum_{n=N_2}^{N_2+p} 2^n\frac{1}{3^nz}-C\right|>|2^{n_0}-\frac{2}{z}-C|>\epsilon.$$
 A: One easily proves the result even uniformly for $z \in K$ compact that doesn't contain $0$ since for all such there is $A=A_K >0, |z| \ge A, z \in K$; then pick $n_K, (8/9)^n \le A$ or equivalently $(2/3)^n \le A(3/4)^n, n \ge n_K$ and use that for $|w| \le 1/2, |\sin w| \le C|w|$ which can be easily seen since $|\sin w /w|$ is bounded and continuos there;
then for $n \ge n_K, z \in K,  |3^nz| \ge A3^n \ge 2$ since $A(9/8)^n \ge 1$ by our choice of $n_K$, hence $|\sin (1/(3^nz)| \le C/|3^nz|$ or $2^n|\sin (1/(3^nz)| \le C(3/4)^n$ again by our choice of $n_K$
So one has uniformly in $z \in K$ that $\sum_{n=1}^\infty |2^n\sin\frac{1}{3^nz}| \le f_K(z)+ \sum_{n \ge n_K}C(3/4)^n$ where $f_K$ is the finite sum of the absolute values of the terms from $1$ ro $n_K-1$ so it is a fixed (for $K$ fixed) continuos function on $K$, so in particularly the OP series converges absolutely and uniformly on $K$
For part $2$ and $z=1/3^N$ the $N$ term of the series is $2^N/\sin 1$ so is unbounded with $N$ which shows that the partial series cannot be uniformly bounded for the sequence $z_N \to 0$, so in particular, they cannot converge uniformly
A: Further to my comment, we can use that $\sin(x)\to x$ for $x\to0$ and look at the ratio of terms:
$$\lim_{n\to\infty}\frac{u_{n+1}}{u_n}=\lim_{n\to\infty}\frac{2^{n+1}\times z\times3^n}{2^n\times z\times 3^{n+1}}=\frac 23<1$$
As you mentioned the problem is as $z\to0$, $\sin\left(\frac 1{3^nz}\right)\not\to0$
